力扣的合併K個排序鍊錶解法（Python3）

題目描述：

合併 k 個排序鍊錶，返回合併後的排序鍊錶。請分析和描述演算法的複雜度。

示例:輸入:

[1->4->5,

1->3->4,

2->6

]輸出: 1->1->2->3->4->4->5->6

# definition for singly-linked list.

# class listnode:
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class
solution
:def
mergeklists
(self, lists: list[listnode])-
> listnode:
result =
for i in lists:
while i :
i  = i.
next
if result ==
:return
result.sort(
)        line1 =  listnode(0)
line2 = line1
while result:
line1.
next
= listnode(result.pop(0)
)            line1 = line1.
next
return line2.
next

參考程式2：

# definition for singly-linked list.

# class listnode:
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class
solution
:def
mergeklists
(self, lists: list[listnode])-
> listnode:
if lists is
none
orlen
(lists)==0
:return
none
while
len(lists)
>1:
temp =
for i in
range(0
,len
(lists)-1
,2):
p = self.mergetwolist(lists[i]
, lists[i +1]
)iflen(lists)%2
==1:-
1])            lists = temp
return lists[0]
defmergetwolist
(self, line1, line2)
:if line1 is
none
:return line2
if line2 is
none
:return line1
result = listnode(0)
r = result
while line1 is
notnone
and line2 is
notnone
:if line1.val <= line2.val:
r.next
= line1
line1 = line1.
next
else
:                r.
next
= line2
line2 = line2.
next
r = r.
next
if line1 is
notnone
:            r.
next
= line1
if line2 is
notnone
:            r.
next
= line2
return result.
next

參考程式3：

# definition for singly-linked list.

# class listnode:
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class
solution
:def
mergeklists
(self, lists: list[listnode])-
> listnode:
import heapq 
result = listnode(-1
)        r = result
line =
list()
temp =
0for i in lists:
while i:
(i.val, temp, i)
)                i = i.
next
temp +=
1while line:
r.next[2
]            r = r.
next
return result.
next

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